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Output waveforms of two diode circuits

Short answer

Figure 1: series resistor and shunt diode clipper gives 0 V positive peak, negative peak −Vm·R2/(R1+R2). Figure 2: series capacitor and shunt diode clamper shifts DC level so output swings from 0 V down to −2Vm, with positive peak clamped at 0 V.

Draw output waveforms of the following circuits and indicate the peak output voltage. Assume diode is ideal. [Figure 1: sine wave Vin, series resistor R1, shunt diode D, load R2, output VOP. Figure 2: sine wave Vin, series capacitor C, shunt diode, load R, output VOP.]

Assumptions: ideal diodes, Vin = Vm sin ωt, and in both circuits the diode anode is on the output side (so it conducts in the positive half cycle).

Figure 1: series R1, shunt diode D, load R2 (clipper)

 Vin o--[R1]--+----+--o VOP
              |    |
              D   R2
              |    |
 GND o--------+----+
  • Positive half cycle: the diode is ON (short circuit) and shorts the load, so VOP = 0.
  • Negative half cycle: the diode is OFF; R1 and R2 form a divider: VOP = Vin × R2/(R1 + R2).

Peak output voltage: VOP(peak) = −Vm R2/(R1 + R2) (negative peak); the positive peak is 0 V.

 VOP
    0 ___________            ___________
                 \          /
                  \        /
 -Vm R2/(R1+R2)    \______/
        (only the negative half appears, reduced by R2/(R1+R2))

Figure 2: series C, shunt diode, load R (clamper)

 Vin o---| |---+---+--o VOP
           C    |   |
                D   R
                |   |
 GND o----------+---+
  • In the first positive quarter cycle the diode conducts and the capacitor charges to Vm (left plate positive).
  • The diode is then OFF, and VOP = Vin − Vm.

Peak output voltage: the positive peak is clamped at 0 V and the negative peak is −2Vm. The waveform is the input sine wave shifted down by Vm.

 VOP
    0 --.--------------.--------------.--
         \            / \            /
          \          /   \          /
           \        /     \        /
 -2Vm       `------'       `------'

If the diode is reversed (cathode at the output), figure 1 clips the negative half and figure 2 shifts the wave up by Vm (the output swings from 0 to +2Vm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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